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May 25, 2015

May 25, 2015

Posted by **jeni** on Tuesday, May 8, 2012 at 11:01pm.

PART A: What is the minimum value of needed to move the two blocks?

PART B: If the force is 10% greater than your answer for (a), what is the acceleration of each block?

- physics- please ! -
**Elena**, Wednesday, May 9, 2012 at 2:12pmThe forces acting on the blocks are:

top block (m1) :

Gravity m1•g (downwards), the normal force on contact between two blocks N1 (upwards), friction force between two blocks F1(fr) ( to the right), Tension T (to the left).

Bottom block (m2).

Gravity m2•g (downwards), gravity m1•g (downwards), the normal force on contact between blocks and the surface N2 (upwards), friction force between two blocks F1(fr) (to the left), friction force between the block and the surface F2(fr) (to the left), tension T (to the left) , unknown force F (to the right).

Equations of the blocks motion (projections on the horizontal and vertical axes):

top block -

x: m1•a = T – F1(fr),

y: 0 = – m1•g + N1. => N1 = m1•g

bottom block –

x: m2•a = F –T –F1(fr) – F2(fr),

y: 0 = - m1•g – m2•g +N2. => N2 = (m1+m2) •g

(a) If the two blocks are just to move, then the force of static friction will be at its maximum, and sothe frictions forces are as follows.

F1(fr) = k1• N1 = k1•m1•g,

F2(fr) = k1• N2 = k1•(m1+m2)•g,

a = 0;

0 = T – F1(fr), => T = F1(fr),

0 = F –T –F1(fr) – F2(fr),

Solve for F.

F = T +F1(fr) + F2(fr) = F1(fr) +F1(fr) + F2(fr) = 2 •F1(fr) + F2(fr) =

= 2• k1•m1•g + k1•(m1+m2)•g = k1•g (3•m1+m2) = 0.65•9.8•(3•3+5) = 89.18 N.

(b) Now force is F1 = 1.1•89.18 = 98.1 N

m1•a = T – F1(fr),

m2•a = F1 –T –F1(fr) – F2(fr),

(m1+m2) •a = F1 - 2 •F1(fr) - F2(fr) =F1 –k2•g•(3•m1+m2),

a = {F1 –k2•g•(3•m1+m2)}/ (m1+m2) =

={98.1 – 0.42•9.8•(3•3+5)}/(3+5) = 5.06 m/s².